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%I #18 Sep 08 2022 08:45:42
%S 288,577,866,1155,1444,1733,2022,2311,2600,2889,3178,3467,3756,4045,
%T 4334,4623,4912,5201,5490,5779,6068,6357,6646,6935,7224,7513,7802,
%U 8091,8380,8669,8958,9247,9536,9825,10114,10403,10692,10981,11270,11559
%N 289n - 1.
%C The identity (289*n-1)^2-(289*n^2-2*n)*(17)^2=1 can be written as a(n)^2-A158252(n)*(17)^2=1.
%H Vincenzo Librandi, <a href="/A158253/b158253.txt">Table of n, a(n) for n = 1..10000</a>
%H Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5785989&tstart=0">X^2-AY^2=1</a>
%H E. J. Barbeau, <a href="http://www.math.toronto.edu/barbeau/home.html">Polynomial Excursions</a>, Chapter 10: <a href="http://www.math.toronto.edu/barbeau/hxpol10.pdf">Diophantine equations</a> (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(17^2*t-2)).
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F a(n) = 2*a(n-1)-a(n-2).
%F G.f.: x*(288+x)/(1-x)^2.
%t 289Range[50]-1 (* _Harvey P. Dale_, Nov 01 2011 *)
%t LinearRecurrence[{2,-1},{288,577},50]
%o (Magma) I:=[288, 577]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]];
%o (PARI) a(n) = 289*n - 1.
%Y Cf. A158252.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 15 2009
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